Problem 3 - Active Calculus Multivariable

Question

Find the partial derivatives of the function $$w=\sqrt{4 r^{2}+3 s^{2}+7 t^{2}}$$ \(\frac{\partial w}{\partial r}=\) ________. \(\frac{\partial w}{\partial s}=\) ________. \(\frac{\partial w}{\partial t}=\) ________.

Step-by-Step Solution

Answer

The short answer based on the step-by-step solution is: \(\frac{\partial w}{\partial r}=\frac{4r}{\sqrt{4r^2 + 3s^2 + 7t^2}}\) \(\frac{\partial w}{\partial s}=\frac{3s}{\sqrt{4r^2 + 3s^2 + 7t^2}}\) \(\frac{\partial w}{\partial t}=\frac{7t}{\sqrt{4r^2 + 3s^2 + 7t^2}}\)

Step 1: Express the given function in the form of \(w^2=\)

First, let's rewrite the given function as a square so that we can later use the chain rule. The given function is: \[w = \sqrt{4r^2 + 3s^2 + 7t^2}\] Squaring both sides, we have: \[w^2 = 4r^2 + 3s^2 + 7t^2\] Now, we will differentiate this equation with respect to r, s, and t.

Step 2: Differentiate with respect to \(r\)

To find the partial derivative with respect to \(r\), we differentiate the equation with respect to \(r\) while keeping \(s\) and \(t\) constant: \[\frac{\partial(w^2)}{\partial r} = \frac{\partial(4r^2 + 3s^2 + 7t^2)}{\partial r}\] Applying the derivative, we get: \[2w\frac{\partial w}{\partial r} = 8r\] Now we need to find the expression for \(\frac{\partial w}{\partial r}\): \[\frac{\partial w}{\partial r}=\frac{8r}{2w}=\frac{4r}{w}\]

Step 3: Differentiate with respect to \(s\)

Now, we will differentiate the equation with respect to \(s\) while keeping \(r\) and \(t\) constant: \[\frac{\partial(w^2)}{\partial s} = \frac{\partial(4r^2 + 3s^2 + 7t^2)}{\partial s}\] Applying the derivative, we get: \[2w\frac{\partial w}{\partial s} = 6s\] Now we need to find the expression for \(\frac{\partial w}{\partial s}\): \[\frac{\partial w}{\partial s}=\frac{6s}{2w}=\frac{3s}{w}\]

Step 4: Differentiate with respect to \(t\)

Lastly, we will differentiate the equation with respect to \(t\) while keeping \(r\) and \(s\) constant: \[\frac{\partial(w^2)}{\partial t} = \frac{\partial(4r^2 + 3s^2 + 7t^2)}{\partial t}\] Applying the derivative, we get: \[2w\frac{\partial w}{\partial t} = 14t\] Now we need to find the expression for \(\frac{\partial w}{\partial t}\): \[\frac{\partial w}{\partial t}=\frac{14t}{2w}=\frac{7t}{w}\]

Step 5: Substitute the expression for \(w\)

Now we can substitute the expression for \(w\) from the original equation \(w = \sqrt{4r^2 + 3s^2 + 7t^2}\) back into our partial derivative expressions: \[\frac{\partial w}{\partial r}= \frac{4r}{\sqrt{4r^2 + 3s^2 + 7t^2}}\] \[\frac{\partial w}{\partial s}= \frac{3s}{\sqrt{4r^2 + 3s^2 + 7t^2}}\] \[\frac{\partial w}{\partial t}= \frac{7t}{\sqrt{4r^2 + 3s^2 + 7t^2}}\] So, the partial derivatives of the function are: \[\frac{\partial w}{\partial r}=\frac{4r}{\sqrt{4r^2 + 3s^2 + 7t^2}}\] \[\frac{\partial w}{\partial s}=\frac{3s}{\sqrt{4r^2 + 3s^2 + 7t^2}}\] \[\frac{\partial w}{\partial t}=\frac{7t}{\sqrt{4r^2 + 3s^2 + 7t^2}}\]

Key Concepts

Multivariable Calculus

In multivariable calculus, we extend the concepts of single-variable calculus to functions that depend on more than one variable. This allows us to study functions in a more comprehensive way. For example, consider a function like \(w = \sqrt{4r^2 + 3s^2 + 7t^2}\), which depends on three variables: \(r\), \(s\), and \(t\).
When working with functions of multiple variables, we can explore how changes in each variable individually affect the function's output, providing insights into the dynamics of the system represented by the function. One fundamental tool in this analysis is the **partial derivative**, which helps determine how a function changes as one specific variable changes while keeping others constant.
This ability to study variations in each direction separately is a cornerstone of multivariable calculus, making it crucial for fields like physics, engineering, and economics.

• The function \(w = \sqrt{4r^2 + 3s^2 + 7t^2}\) represents a surface in \(3D\) space.
• Understanding its behavior involves calculating partial derivatives with respect to each variable.

Chain Rule

The **chain rule** in calculus is a powerful method for finding the derivative of composite functions. In its essence, the chain rule helps us differentiate functions where one function is nested inside another. The chain rule is especially beneficial in multivariable calculus, where functions frequently depend on multiple other functions.
For example, in our problem, the function \(w = \sqrt{4r^2 + 3s^2 + 7t^2}\) can be thought of as a function \(w\) of another function \(u = 4r^2 + 3s^2 + 7t^2\). The derivative of \(w\) with respect to any of its variables, like \(r\), can be found using the chain rule:

• First, differentiate \(u\) with respect to \(r\), yielding partial derivatives of \(4r^2\).
• Then multiply the result by the derivative of the outer function, \(\sqrt{u}\), with respect to \(u\).

This technique enables the differentiation of more complex forms systematically and accurately.

Differentiation

**Differentiation** is the process of finding the derivative of a function, which measures the rate at which a function is changing at any given point. Partial differentiation specifically refers to this process applied to multivariable functions.
Consider the function \(w = \sqrt{4r^2 + 3s^2 + 7t^2}\). To differentiate this function with respect to one of its variables, such as \(r\), we treat all other variables as constants. This focuses on understanding how changes in \(r\) alone affect \(w\).
The procedure involves several key steps:

• Start by expressing \(w\) in a simpler form if possible, such as \(w^2 = 4r^2 + 3s^2 + 7t^2\).
• Apply the differentiation rules to isolate the desired partial derivative.
• For instance, differentiating \(4r^2\) with respect to \(r\) gives \(8r\).

This allows us to derive expressions like \(\frac{\partial w}{\partial r}\), providing a clear picture of how each individual variable functions in the composition of \(w\). This clarity in understanding is fundamental for solving real-world problems involving changes along multiple dimensions.